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Section One Precalc Review, Average Value, Exponential Growth & Decay, and Slope Fields By: RE, Rusty, Matthew, and D Money.

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Presentation on theme: "Section One Precalc Review, Average Value, Exponential Growth & Decay, and Slope Fields By: RE, Rusty, Matthew, and D Money."— Presentation transcript:

1 Section One Precalc Review, Average Value, Exponential Growth & Decay, and Slope Fields By: RE, Rusty, Matthew, and D Money

2 Unit Circle Geometrical figure relating trig functions to specific angles and coordinates

3 Unit Circle Angle is in radians and measured counterclockwise from 0 radians. Coordinates are always fractions of 1; 1 is the radius of the unit circle

4 Practice Problem 1 Graph the following equation using only skills from pre-calc: Y=4/(x^2-9) Vertical asymptotes are found by setting denominator equal to 0, then solving for x (x^2-9)=0 (x+3)(x-3)=0 X=3,-3

5 Practice Problem 1 Y=4/(x^2-9) Vertical asymptotes at x=3,-3 Parent graph of 1/(x^2)  – Stretched vertically fourfold Test values on either side of each asymptote X=0, y=?; x=-4, y=?; x=4, y=? (0, -4/9) negative; (-4, 4/5) positive; (4, 4/5) positive

6 Solution 1

7 Practice Problem 2 Ratio of donkeys to people in Kazakhstan – 1:3 Donkey population of Kazakhstan in 1900 – 7 Growth rate of people in Kazakhstan -.23857 How many donkeys were in Kazakhstan in 1999? – Assume that the donkey-people ratio remains constant, as it generally does

8 Practice Problem 2 Determine human population of Kazakhstan in 1900 – Donkeys/People=1/3 – 7/P=1/3 – P=21 people Use growth equation with variables – y=Ye^kt – y is people in 1999; Y is people in 1900; k is growth rate of people; t is time in years

9 Practice Problem 2 y=(21)e^(.23857*99) y=379,815,876,900 people – They reproduce rather quickly How many donkeys? – Donkeys/People=1/3 – D/379,815,876,900=1/3 D=126,605,292,300 donkeys in Kazakhstan in 1999

10 126,605,292,300 DONKEYS!!!

11 Practice Problem 3 Find the average value of f(x) on the interval [7,20] Given:

12 Practice Problem 3 Find the value of the integral of f(x) from 7 to 20 in terms of known values

13 Practice Problem 3 Find the value of the integral of f(x) from 7 to 20 in terms of known values =-(-17)-(-12)=29

14 Practice Problem 3 Use average value formula: Plug in variables: Now, solve: =1/13(29)= 29/13

15 Practice Problem 4 Graph the slope field of the differential equation dy/dx=y-x Make a table of values X:0000000 Y:0123-1-2-3 Dy/dx:0123-1-2-3

16 Practice Problem 4 X:1111111 Y:0123-1-2-3 Dy/dx:-1012-2-3-4 X:2222222 Y:0123-1-2-3 Dy/dx:-2-101-3-4-5 X:3333333 Y:0123-1-2-3 Dy/dx:-3-2-10-4-5-6

17 Practice Problem 4 X:-1-1-1-1-1-1-1 Y:0123-1-2-3 Dy/dx:12340-1-2 X:-2-2-2-2-2-2-2 Y:0123-1-2-3 Dy/dx:234510-1 X:-3-3-3-3-3-3-3 Y:0123-1-2-3 Dy/dx:3456210

18 Practice Problem 4 Plotting all points with slope dy/dx on a graph with domain and range both of [-3,3] should look like this:


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